To solve this problem, we can break it down into three distinct phases:

1. Acceleration phase: The car accelerates from rest.
2. Constant speed phase: The car moves at a constant speed.
3. Deceleration phase: The car slows down uniformly to a stop.

Let's go through each part step by step.

(a) Total Time in Motion

Phase 1: Acceleration

• Initial velocity: $u_1 = 0 \, \text{m/s}$
• Final velocity: $v_1 = 34.0 \, \text{m/s}$
• Acceleration: $a_1 = 2.00 \, \text{m/s}^2$

We use the first kinematic equation to find the time taken to reach the speed of 34.0 m/s: $v_1 = u_1 + a_1 t_1$ Substituting the known values: $34.0 = 0 + 2.00 t_1$ $t_1 = \frac{34.0}{2.00} = 17.0 \, \text{s}$

Phase 2: Constant Speed

• Time at constant speed: $t_2 = 37.0 \, \text{s}$

Phase 3: Deceleration

• Initial velocity: $u_3 = 34.0 \, \text{m/s}$
• Final velocity: $v_3 = 0 \, \text{m/s}$
• Time to stop: $t_3 = 5.00 \, \text{s}$

We use the same kinematic equation as in Phase 1 to find the acceleration: $v_3 = u_3 + a_3 t_3$ Here, $v_3 = 0$ (since the car stops), so: $0 = 34.0 + a_3 \times 5.00$ $a_3 = \frac{-34.0}{5.00} = -6.80 \, \text{m/s}^2$ The deceleration is $-6.80 \, \text{m/s}^2$, but the time is already given as 5.00 s.

Now, sum up the times from all three phases: $\text{Total time} = t_1 + t_2 + t_3 = 17.0 \, \text{s} + 37.0 \, \text{s} + 5.00 \, \text{s} = 59.0 \, \text{s}$

Final Answer for (a):

(a) The ride-sharing car is in motion for 59.0 seconds.

(b) Average Velocity

The average velocity is calculated as the total displacement divided by the total time in motion.

Phase 1: Displacement during Acceleration

Use the kinematic equation: $d_1 = u_1 t_1 + \frac{1}{2} a_1 t_1^2$ Substituting the known values: $d_1 = 0 \times 17.0 + \frac{1}{2} \times 2.00 \times (17.0)^2 = \frac{1}{2} \times 2.00 \times 289 = 289.0 \, \text{m}$

Phase 2: Displacement during Constant Speed

Since the car is moving at a constant speed: $d_2 = v_1 \times t_2 = 34.0 \times 37.0 = 1258.0 \, \text{m}$

Phase 3: Displacement during Deceleration

Use the kinematic equation again: $d_3 = u_3 t_3 + \frac{1}{2} a_3 t_3^2$ Substituting the known values: $d_3 = 34.0 \times 5.00 + \frac{1}{2} \times (-6.80) \times (5.00)^2$ $d_3 = 170.0 - \frac{1}{2} \times 6.80 \times 25.0 = 170.0 - 85.0 = 85.0 \, \text{m}$

Total Displacement

$\text{Total displacement} = d_1 + d_2 + d_3 = 289.0 + 1258.0 + 85.0 = 1632.0 \, \text{m}$

Average Velocity

$\text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}} = \frac{1632.0 \, \text{m}}{59.0 \, \text{s}} \approx 27.66 \, \text{m/s}$

Final Answer for (b):

(b) The average velocity of the ride-sharing car is 27.66 m/s.

Would you like more details or have any questions about the solution?

Here are some related questions you might find useful:

1. What would be the final velocity if the car decelerated over a longer period, say 7 seconds?
2. How much time would the car spend in motion if the initial acceleration was higher, like 3 m/s²?
3. What would be the total displacement if the constant speed phase lasted 50 seconds instead of 37 seconds?
4. How would the average velocity change if the car reached a higher final speed of 40 m/s?
5. What if the car was moving uphill during the constant speed phase, requiring additional force? How would this affect the results?

Tip: When calculating average velocity over multiple phases of motion, always ensure you account for both time and displacement accurately in each phase.